## 8:00 AM Class Group 4

Meat and Potatoes Group Project

Group 4: Chris Cardenas, Chase Chemero, Joseph Ciarleglio, Jennifer Rosa, Thanh Tran

Math 1070Q

Linear Programming Project

December 1, 2008

### Group Project Meat and Potatoes

We first assign the number of portions ordered of potatoes and meat 'X' and 'Y' respectively. One portion of potatoes (1X) contains 3 units of carbohydrates, 4 units of vitamins, 1 unit of protein and costs 75 cents. One portion of meat (1Y) contains 1 unit of carbohydrates, 3 units of vitamins, 3 units of protein and costs 2 dollars. Because we know there can only be positive numbers of portions ordered we can infer non-negativity clauses (X ≥0 and Y ≥0). The minimum daily requirements of a balanced diet are 8 units of carbohydrates, 19 units of vitamins and 7 units of protein. These give us our constraints (or conditions).

3X+Y ≥8 ----This constraint that defines the minimum amount of carbohydrates we need daily. One portion of potatoes delivers 3 units of carbohydrates and one portion of meat delivers only one unit of carbohydrates. We need 8 total units of carbohydrates altogether so 3X+Y ≥ 8 (3 times the number of portions ordered of potatoes plus 1 times the number of portions ordered of meat is greater or equal to 8 total units of carbohydrates).

4X+3Y ≥19 -------This constraint defines the minimum amount of vitamins we need daily. One portion of potatoes delivers 4 units of vitamins and one portion of meat delivers 3 units of vitamins. We need 19 total units of vitamins altogether so 4X+3Y ≥19 (4 times the number of portions ordered of potatoes plus 3 times the number of portions ordered of meat is greater or equal to 19 total units of vitamins).

X+3Y ≥7 ------This constraint defines the minimum amount of protein we need daily. One portion of potatoes delivers 1 unit of protein and one unit of meat delivers 3 units of protein. We need 7 units of protein altogether so X+3Y ≥7 (1 times the number of portions ordered of potatoes plus 3 times the number of portions ordered of meat is greater than or equal to 7 total units of protein).

Because we know the costs of both X and Y we can derive our objective function: .75X+2Y=M

Our objective function is crucial to our model because it is what we seek to minimize. M is the cost when ordering X portions of potatoes and Y portions of meat. By minimizing M and still meeting minimum daily requirements we have answered this problem.